This well has a $\psi$ which we can combine with $\psi_I$, $\psi_{II}$ and $\psi_{III}$. I have been playing around and got expressions for them, but they are not the same for ODD and EVEN solutions but lets do this only for ODD ones.

I first used constants above to again draw a graph of transcendental equation and i found 2 possible energies $W$ (i think those aren't quite accurate but should do). This looks like in any QM book (thanks to @Chris White):

Lets chose only one of the possible energies and try to plot $\psi_I$, $\psi_{II}$ and $\psi_{III}$. I choose energy which is equal to $0.17\, W_p$ and calculate constants $\mathcal K$ and $\mathcal L$:

Now when the picture above looks like in a book i will try to use constants $\mathcal K$, $\mathcal L$ and $\boxed{A \!=\! 1}$ (like @Chris White sugested) to plot $\psi_I$, $\psi_{II}$ and $\psi_{III}$. Even now the boundary conditions at $-\tfrac{d}{2}$ and $\tfrac{d}{2}$ are not met:

EXTENDED QUESTION:

It looks like boundary conditions are not met. I did calculate my constants quite accurately, but i really can't read the energies (which are graphicall solutions to the first graph) very accurately. Does anyone have any suggestions on how to meet the boundary conditions?

So do i have to first get allowed energies $W$ from graphical solution of transcendental equations and use them to find possible $\mathcal L$ an $\mathcal{K}$ and finally calculating $A$ with normalisation. Then finally plotting the equation?
–
71GAApr 2 '13 at 7:38

1

Yes. The boundary conditions on $\psi$ can only be satisfied for values of $W$ that satisfy the quantization condition (the transcendental equation).
–
Michael BrownApr 2 '13 at 8:52

You have the wrong sign inside the square root in $\mathcal{K}$. Make that change, and you will find exactly 2 bound solutions ($W<W_p$). As it stands currently, the only real solutions for $(\mathcal{L},\mathcal{K})$ correspond to $W>W_p$, but in this case you know the solution in regions I and III will be sines and cosines rather than exponentials, but such solutions cannot be accommodated in the form currently written.
–
Chris WhiteApr 2 '13 at 23:28

2 Answers
2

Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator:
$$ \hat{H} \psi = E \psi. $$
As such, we expect the solutions to be determined only up to scaling. Clearly if $\psi_n$ is a solution with eigenvalue $E_n$, then
$$ \hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n \psi_n = E_n (A \psi_n) $$
for any constant $A$, so $\psi$ can always be rescaled. In this sense, there is no physical meaning associated with $A$.

To actually choose a value, for instance for plotting, you need some sort of normalization scheme. For square-integrable functions, we often enforce
$$ \int \psi^* \psi = 1 $$
in order to bring the wavefunction more in line with the traditional definition of probability (which says the sum of probabilities is $1$, also an arbitrary constant).

If you happen to be in the regime $E > W_p$, then $\mathcal{K}$ will be imaginary, $\psi_\mathrm{I}$ and $\psi_\mathrm{III}$ will be oscillatory rather than decaying, and the first and third of those integrals will not converge. You could pick an $A$ that conforms to some sort of "normalizing to a delta function," but there are many different variations on this, especially for a split-up domain like this. In that case I would recommend picking an $A$, if you really have to do it, based on some other criterion, such as $\max(\lvert \psi \rvert) = 1$ or something.

I found the case myself. There was a mistake in a GNUPLOT script. The line:

g(x) = -( A*exp(-L*(d/2)) )/( sin(L*(d/2)) )*sin(L*x)

Should have $\mathcal K$ in place of the first $\mathcal L$. This was the first mistake but after i fixed it my graphs still were sloppy, so i redid all the readings for energies from a graphicall solutions to the transcendental equations and recalculated $\mathcal K$ and $\mathcal L$. As it turns out my graphs now came out perfectly!